Prove the least squares approximation property of Legendre polynomials [see (9.5) and (9.6)] as follows. Let f(x) be the given function to be approximated. Let the functions p_l(x)be the normalized Legendre polynomials, that is, p_l(x) = √(2l+1)/2 P_l(x) , so that

∫_-1¹[p_l(x)"]"²dx=1.

Show thatLegendre series for f(x)as far as the p₂(x)term is

f(x)=c₀p₀(x)+c₁p₁(x) +c₃p₃(x) with c₁ =∫_-1¹f(x)p_l(x) dx

Write the quadratic polynomial satisfying the least squares condition as b₀p₀(x)+b₁p₁(x)+b₀p₀(x)by Problem 5.14 any quadratic polynomial can be written in this form). The problem is to find b₀, b₁, b₂so that I=∫_-1¹[f²(x)+(b₀-c₀)²+(b₁-c₁)²+(b₂-c₂)² -c₀² -c₁² -c₂²] dx

Now determine the values of the b's to make I as small as possible. (Hint: The smallest value the square of a real number can have is zero.) Generalize the proof to polynomials of degree n.

The initial conditions that are given for unknown variable and least square approximation of given Legendre polynomials expression,

p_l(x) =√(2l+1)/2 p_l(x)

With their normalized polynomials,

∫_-1¹p_l(x)² dx=1

And here Legendre series is, f(x)=c₀p₀(x)+c₁p₁(x)+c₂p₂(x).

Where constant is,

c₁=∫_-1¹f(x) p₁(x)dx

Here quadratic polynomial in terms of b₀, b₁, b₂is shown below:

I=∫_-1¹[f(x) - (b₀p₀(x)+b₁p₁(x)+b₂p₂(x))]² dx

With sum of the integrals using individual terms is shown below:

I=∫_-1¹[f²(x) - (b₀-c₀)²+(b₁- p₁)²+(b₂-p₂)² -c₀² -c₁² -c₂² dx]

Here the least squares approximation property of legendry polynomials as shown below:

p_l(x)=√(2l+1)/2 p_l(x)

Let function to be approximated and P_l(x) denoted as normalized legendry polynomials

With their normalized polynomials:

∫_-1¹p_l(x)²dx=∫_-1¹ [√(2l+1)/2 p_l(x)]²dx=1

First the constant term is assumption considering c₀, c₁, c₂ that having the function values c₁=∫_-1¹f(x)p₁(x) dx of constant the Legendre series for f(x) as far as the P₂(x) term is ,

Where constant is

c_n=∫_-1¹f(x) p_n(x) dx

Where n=0, 1, 2.

d/dx ((x^-αJ_α(x)) = J_α-1(x) -J_α+1(x)

d/dx((x^-αJ_α(x)) = ∑_j=1^∞[(-1)^j 2j / j!T (j+α+1) x^2j-1 / 2^2j+α]

d/dx((x^-αJ_α(x)) = ∑_j=1^∞[(-1)^j 2j / (j-1) !T (j+α+1) x^2j-1 / 2^2j-1+α]

d/dx((x^-αJ_α(x)) = ∑_j=1^∞[(-1)^j 2j / k !T (k+1+(α+1)) (x/2)^(2k+1+α)]

Therefore,

d/dx((x^-αJ_α(x)) = -x^-αJ_α+1(x)

Similarly function f(x)=c₀p₀(x)+c₁p₁(x)+c₂p₂(x) with sum of integral ∫_-1¹p_l(x)²dx=1 becomes the quadratic polynomial which satisfied the least squares condition as,

I=∫_-1¹[f(x) - (b₀p₀(x)+(b₁p₁(x)+(b₂p₂(x)]²dx b₀p₀(x) +b₁c₁(x) +b₂c₂(x)

d/dx J_α(x) = d/dx (x^α(x^-αJ_α(x)) = αx^α-1 (J_α(x)) + x^α d/dx (x^αJ_α(x))

d/dx J_α(x) = -αx^α-1J_α(x) + x^-α(x^αJ_α-1(x))

d/dx J_α(x) = -αx^α-1J_α(x) +J_{α -1}(x)

Any quadratic polynomial written as and found, b₀, b₁, b₂ values.

Here, quadratic polynomial in terms of b₀, b₁, b₂ is used in this expression shown below.

I=∫_-1¹[f(x) - (b₀p₀(x)+b₁p₁(x)+b₂p₂(x)]² dx

By squaring and rearranging found:

I=∫_-1¹[f²(x) - (b₀-c₀)²+(b₁- p₁)²+(b₂-p₂)² -c₀² -c₁² -c₂² dx]

Further this expression used w.r.t. sum of the integrals using individual terms is shown below.

I=∫_-1¹[f²(x) - (b₀-c₀)²+(b₁- p₁)²+(b₂-p₂)² -c₀² -c₁² -c₂² dx]